Question

A compound contains ${10^{ - 2}}\% $ of phosphorus. If the atomic mass of $P$ is $31$, then the molecular mass of the compound having one phosphorus atom per molecule is:
A. $3.1 \times {10^5}$
B. $3.1$
C. $31$
D. None of these

Answer 1

Hint:Phosphorus is a chemical element with the symbol (\[P\] ) and atomic number 15. Elemental phosphorus exists in two major forms which are white phosphorus and red phosphorus, but because it is highly reactive, phosphorus is never found as a free element on Earth. It has a concentration in the Earth's crust of about one gram per kilogram (compare copper at about \[0.06\] grams). In minerals, phosphorus generally occurs in the form of phosphate.

Complete step by step answer:
As per the question, a particular type of compound contains ${10^{ - 2}}\% $ of phosphorus. This can be represented mathematically as:
${10^{ - 2}}\% $ of phosphorous = $\dfrac{{{{10}^{ - 2}}}}{{100}} = \dfrac{1}{{{{10}^4}}}$
The above statement means that for every ${10^4}$ parts of the compound, the part of phosphorus is 1.
Let us assume the molecular mass of the compound to be $x$ .
The amount of phosphorus present in the compound will be equal to the molecular mass of phosphorus. Mathematically, this can be written as:
$ \Rightarrow {10^{ - 2}}\% $ of $x = 31$
Hence, $\dfrac{{{{10}^{ - 2}}}}{{100}} \times x = 31$
On solving the above equation, we have:
$ \Rightarrow x = 31 \times {10^4} = 3.1 \times {10^5}$
This means that If the atomic mass of $P$ is $31$, then the molecular mass of the compound having one phosphorus atom per molecule is $3.1 \times {10^5}$.
Thus, the correct option is A. $3.1 \times {10^5}$ .


Note:
Phosphorus is essential for life. Phosphates (compounds containing the phosphate ion,$PO_4^{3 - }$ ) are a component of DNA, RNA, ATP, and phospholipids. Elemental phosphorus was first isolated from human urine, and bone ash was an important early phosphate source. Phosphate mines contain fossils because phosphate is present in the fossilized deposits of animal remains and excreta.